Diodes and grounds

Thread Starter

JoyAm

Joined Aug 21, 2014
126
How can input B be at 3 volts when it is connected to a source of 2 volts?

Since all the connections to the bottom of the resistor are made along one wire they must be at the same voltage, which we have agreed is 1 volt, because the diode input C must be on.
So there should be no questions marks.

Both the diodes A and B are off, so isolate the 3volts and 2volts at their cathodes, from the 1 volt at their common anode connection with the resistor.

Yeah i meant to put 2 on B, stupid typo i guess . I think i understand now, thanks :)

I think first you should redraw it like this:View attachment 72715
And then this:View attachment 72716
These drawings are lovely and very helpful, thanks kubeek
 

studiot

Joined Nov 9, 2007
4,998
This sketch may help, it is an amplification of post#15.

Let us start by making sure all the diodes are off.
I have done that by connecting them to the 5 volt supply through switches in Fig1.

Now for the rest I have dropped input B since it is only a distraction.

I have two choices, switch input A or switch input C to the set voltage first.

I have show this in fig2 and fig3.
The resulting voltages are shown and it can be seen that diode A is on and diode C is off in fig 2
but diode C is on and Diode A is off in fig3.

If I now switch the second input to its set voltage the result is the same diagram fig4, with diode A off and diode C on.

So it does not matter how I arrive at the set voltages, the result is the same.

diodeor2.jpg
 

studiot

Joined Nov 9, 2007
4,998
Kubeek, look again at your redrawing.

You have grounded the third diode cathode, rather than connected it to the set voltage.
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
This sketch may help, it is an amplification of post#15.

Let us start by making sure all the diodes are off.
I have done that by connecting them to the 5 volt supply through switches in Fig1.

Now for the rest I have dropped input B since it is only a distraction.

I have two choices, switch input A or switch input C to the set voltage first.

I have show this in fig2 and fig3.
The resulting voltages are shown and it can be seen that diode A is on and diode C is off in fig 2
but diode C is on and Diode A is off in fig3.

If I now switch the second input to its set voltage the result is the same diagram fig4, with diode A off and diode C on.

So it does not matter how I arrive at the set voltages, the result is the same.

View attachment 72717
The figure 3 is has same results figure 4 because Vc<Va ? so if Va was for example 0.9V diode A would be on and C off thus the output voltage would be 0.9?
Also from what i see here the output voltage ( at the F point of your first draw ) is equal with the voltage at the anode of the diode that is turned on, If that is true what would happen if in another circuit more than one diodes were on ? would we sum their voltages up ?
 

studiot

Joined Nov 9, 2007
4,998
In principle, yes if Va<Vc (Va=0/9; Vc=1) then A would be on and C would be off.
However 0.1 volts difference would not be reliable.

Remember I said this circuit was developed for 0.2 volt germanium diodes and 10 volt signals.
With a 5 volt supply and modern silicon diodes the diode forward drop needs to be taken into account as 0.6 to 0.9 volts are a significantly greater proportion of 5 than 0.2 is of 10.

You can now play about with a few configurations for yourself.
That is the best way to become familiar.

These circuits needed to higher voltages to operate anywhere near reliably and could not accept concatenation of many stages so were soon dropped for better circuits using transistors.

Today the configuration is used in control and bus circuits as a single stage and are called steering diodes (look these up). They are also used within memory chips and LSI
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
In principle, yes if Va<Vc (Va=0/9; Vc=1) then A would be on and C would be off.
However 0.1 volts difference would not be reliable.

Remember I said this circuit was developed for 0.2 volt germanium diodes and 10 volt signals.
With a 5 volt supply and modern silicon diodes the diode forward drop needs to be taken into account as 0.6 to 0.9 volts are a significantly greater proportion of 5 than 0.2 is of 10.

You can now play about with a few configurations for yourself.
That is the best way to become familiar.

These circuits needed to higher voltages to operate anywhere near reliably and could not accept concatenation of many stages so were soon dropped for better circuits using transistors.

Today the configuration is used in control and bus circuits as a single stage and are called steering diodes (look these up). They are also used within memory chips and LSI
Thanks :)
Is my assumption about the voltage output right ?
 

studiot

Joined Nov 9, 2007
4,998
Also from what i see here the output voltage ( at the F point of your first draw ) is equal with the voltage at the anode of the diode that is turned on,
Yes this is true if we ignore diode voltage drops.

If that is true what would happen if in another circuit more than one diodes were on ? would we sum their voltages up ?
Why would you think that? You sum voltages that are in series.

Do you not think it is a good idea for you to play with a pencil and paper as I suggested and try out a few ideas for yourself?
 

Thread Starter

JoyAm

Joined Aug 21, 2014
126
Yes this is true if we ignore diode voltage drops.



Why would you think that? You sum voltages that are in series.

Do you not think it is a good idea for you to play with a pencil and paper as I suggested and try out a few ideas for yourself?
Yes i think i should practice a bit, thank you :)
 

studiot

Joined Nov 9, 2007
4,998
Hello, Scott,

Please read carefully my various posts.
The circuit I posted is an OR gate implemented in diode-resistor logic with V=10 volts and input signals of 0volts= 0 and -10volts=1.
The output will be -10 volts if any one input is -10V.
The transfer function is

F=A+B+C

Which is the correct transfer function for an OR gate.

Pleae note I was writing my pic whilst the OP was posting further information, so I did not have the 5V and ground.
 

ScottWang

Joined Aug 23, 2012
7,397
Hello, Scott,

Please read carefully my various posts.
The circuit I posted is an OR gate implemented in diode-resistor logic with V=10 volts and input signals of 0volts= 0 and -10volts=1.
The output will be -10 volts if any one input is -10V.
The transfer function is

F=A+B+C

Which is the correct transfer function for an OR gate.

Pleae note I was writing my pic whilst the OP was posting further information, so I did not have the 5V and ground.
Yes, I already review your other posts, but that were you did something as tone jumping, the op was talking about the positive logic, but you were talking about the negative logic and you didn't label any voltage on the inputs, who will know that just from that circuit?
 

studiot

Joined Nov 9, 2007
4,998
Yes, I already review your other posts, but that were you did something as tone jumping, the op was talking about the positive logic, but you were talking about the negative logic and you didn't label any voltage on the inputs, who will know that just from that circuit?
But clearly not carefully enough.

but that were you did something as tone jumping, the op was talking about the positive logic, but you were talking about the negative logic ?
I did not have the benefit of knowing positive or negative logic when I posted, the only posts I had to read were posts#1 and 2, or even if that was what the OP was asking about.

In fact I think it pretty remarkable that my guess as to what post#1 might be about came so close in answer to

Hello there, i just started studying microelectronics after getting some basic knowledge of electrical circuits and i find myself terribly confused. In circuits with diodes i see no meshes,
By all means add to this thread, even at this late stage.
I would particularly like to hear your views on the type of logic that accepts +1V ; +2V and +3V as valid inputs, with a +5V supply and ground.
 

ScottWang

Joined Aug 23, 2012
7,397
In fact I think it pretty remarkable that my guess as to what post#1 might be about came so close in answer to
Oh, yeah, you are a Prophet, good for you, you got the good sixth sense ... :)

By all means add to this thread, even at this late stage.
I would particularly like to hear your views on the type of logic that accepts +1V ; +2V and +3V as valid inputs, with a +5V supply and ground.
Assuming the Vf of diode is equal to 0.7V, for a TTL positive logic, the TTL high level is 2.4V.

if anyone of input is low then the output will be low, so this function is an AND gate.

If the +1V input is high(1V), then the output voltage will be like Vo=1V+0.7V=1.7V, if the +2V is high(2V), 2V>1.7V, the diode will be off, the +3V also the same, the diode is off, so the Vo=1.7V, now, from the calculation that we can see the circuit doesn't like the normal AND gate.

In another word, the +2V, +3V seem useless, what purpose is this circuit?
Code:
+1V   +2V   +3V    Vo
==========================
Lo     x     x    LO(0.7V)
x     Lo     x    LO(0.7V)
x      x     Lo   LO(0.7V)
---------------------------------------------
1V    2V     3V    1V+0.7V
 
Last edited:

studiot

Joined Nov 9, 2007
4,998
In another word, the +2V, +3V seem useless, what purpose is this circuit?
I don't know it is not my circuit.

But the other use of diodes I mentioned backalong was steering diodes. They are not restricted to standard logic levels.
 

ScottWang

Joined Aug 23, 2012
7,397
I don't know it is not my circuit.

But the other use of diodes I mentioned backalong was steering diodes. They are not restricted to standard logic levels.
Yes, I knew that wasn't your circuit.
If the inputs are not like as 0V,1V and 0V,2V and 0V,3V, and they are like as floating,1V and floating,2V and floating,3V , then they can be used as priority circuit, the +1V is the most priority, the +3Vis the least priority.
 
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